Open Access

Some properties of meromorphically multivalent functions

Journal of Inequalities and Applications20122012:86

https://doi.org/10.1186/1029-242X-2012-86

Received: 18 October 2011

Accepted: 16 April 2012

Published: 16 April 2012

Abstract

By using the method of differential subordinations, we derive certain properties of meromorphically multivalent functions.

2010 Mathematics Subject Classification: 30C45; 30C55.

Keywords

analytic function meromorphically multivalent function subordination

1 Introduction

Let Σ(p) denotes the class of meromorphically multivalent functions f(z) of the form
f z = z - p + k = 1 a k - p z k - p p N = 1 , 2 , 3 , ,
(1.1)
which are analytic in the punctured unit disk
U * = z : z C and 0 < z < 1 = U \ 0 .

Let f(z) and g(z) be analytic in U. Then, we say that f(z) is subordinate to g(z) in U, written f(z) g(z), if there exists an analytic function w(z) in U, such that |w(z)| ≤ |z| and f(z) = g(w(z)) (z U). If g(z) is univalent in U, then the subordination f(z) g(z) is equivalent to f(0) = g(0) and f(U) g(U).

Let p(z) = 1 + p1z + ... be analytic in U. Then for -1 ≤ B < A ≤ 1, it is clear that
p z 1 + A z 1 + B z z U
(1.2)
if and only if
p z - 1 - A B 1 - B 2 < A - B 1 - B 2 - 1 < B < A 1 ; z U
(1.3)
and
Re p z > 1 - A 2 B = - 1 ; z U .
(1.4)

Recently, several authors (see, e.g., [17]) considered some interesting properties of meromorphically multivalent functions. In the present article, we aim at proving some subordination properties for the class Σ(p).

To derive our results, we need the following lemmas.

Lemma 1 (see [8]. Let h(z) be analytic and starlike univalent in U with h(0) = 0. If g(z) is analytic in U and zg'(z) h(z), then
g z g 0 + 0 z h t t d t .
Lemma 2 (see [9]. Let p(z) be analytic and nonconstant in U with p(0) = 1. If 0 < | z0 | < 1 and Re p z 0 = min z z 0 Re p z , then
z 0 p z 0 - 1 - p z 0 2 2 1 - Re p z 0 .

2 Main results

Our first result is contained in the following.

Theorem 1. Let α ( 0 , 1 2 ] and β (0,1). If f(z) Σ(p) satisfies f(z) ≠ 0 (z U*) and
z - p f z z f z f z + p < δ z U ,
(2.1)
where δ is the minimum positive root of the equation
α sin π β 2 x 2 - x + 1 - α sin π β 2 = 0 ,
(2.2)
then
arg f z z - p - α < π 2 β z U .
(2.3)

The bound β is the best possible for each α ( 0 , 1 2 ] .

Proof. Let
g x = α sin π β 2 x 2 - x + 1 - α sin π β 2 .
(2.4)
We can see that the Equation (2.2) has two positive roots. Since g(0) > 0 and g(1) < 0, we have
0 < α 1 - α δ δ < 1 .
(2.5)
Put
f z z - p = α + 1 - α p z .
(2.6)
Then from the assumption of the theorem, we see that p(z) is analytic in U with p(0) = 1 and α + (1 - α)p(z) ≠ 0 for all z U. Taking the logarithmic differentiations in both sides of (2.6), we get
z f z f z + p = 1 - α z p z α + 1 - α p z
(2.7)
and
z - p f z z f z f z + p = 1 - α z p z α + 1 - α p z 2
(2.8)
for all z U. Thus the inequality (2.1) is equivalent to
1 - α z p z α + 1 - α p z 2 δ z .
(2.9)
By using Lemma 1, (2.9) leads to
0 z 1 - α p t α + 1 - α p t 2 d t δ z
or to
1 - 1 α + 1 - a p z δ z .
(2.10)
In view of (2.5), (2.10) can be written as
p z 1 + α 1 - α δ z 1 - δ z .
(2.11)
Now by taking A = α 1 - α δ and B = -δ in (1.2) and (1.3), we have
arg f z z - p - α = arg p z < arcsin δ 1 - α + α δ 2 = π 2 β

for all z U because of g(δ) = 0. This proves (2.3).

Next, we consider the function f(z) defined by
f z = z - p 1 - δ z z U * .
It is easy to see that
z - p f z z f z f z + p = δ z < δ z U .
Since
f z z - p - α = 1 - α 1 + α 1 - α δ z 1 - δ z ,
it follows from (1.3) that
sup z U arg f z z - p - α = arcsin δ 1 - α + α δ 2 = π 2 β .

Hence, we conclude that the bound β is the best possible for each α ( 0 , 1 2 ] .

Next, we derive the following.

Theorem 2. If f(z) Σ(p) satisfies f(z) ≠ 0 (z U*) and
Re z - p f z z f z f z + p < γ z U ,
(2.12)
where
0 < γ < 1 2 log 2 ,
(2.13)
then
Re z - p f z > 1 - 2 γ log 2 z U .
(2.14)

The bound in (2.14) is sharp.

Proof. Let
p z = f z z - p .
(2.15)
Then p(z) is analytic in U with p(0) = 1 and p(z) ≠ 0 for z U. In view of (2.15) and (2.12), we have
1 - z p z γ p 2 z 1 + z 1 - z ,
i.e.,
z 1 p z 2 γ z 1 - z .
Now by using Lemma 1, we obtain
1 p z 1 - 2 γ log 1 - z .
(2.16)
Since the function 1 - 2γ log(1 - z) is convex univalent in U and
Re 1 - 2 γ log 1 - z > 1 - 2 γ log 2 z U ,

from (2.16), we get the inequality (2.14).

To show that the bound in (2.14) cannot be increased, we consider
f z = z - p 1 - 2 γ log 1 - z z U * .
It is easy to verify that the function f(z) satisfies the inequality (2.12). On the other hand, we have
Re z - p f z 1 - 2 γ log 2

as z → -1. Now the proof of the theorem is complete.

Finally, we discuss the following theorem.

Theorem 3. Let f(z) Σ(p) with f(z) ≠ 0 (z U*). If
Im z f z f z f z z - p - λ < λ λ + 2 p z U
(2.17)
for some λ(λ > 0), then
Re f z z - p > 0 z U .
(2.18)
Proof. Let us define the analytic function p(z) in U by
f z z - p = p z .
Then p(0) = 1, p(z) ≠ 0 (z U) and
z f z f z f z z - p - λ = p z - λ z p z p z - p z U .
(2.19)
Suppose that there exists a point z0(0 < | z0 | < 1) such that
Re p z > 0 z < z 0 and p z 0 = i β ,
(2.20)
where β is real and β ≠ 0. Then, applying Lemma 2, we get
z 0 p z 0 - 1 + β 2 2 .
(2.21)
Thus it follows from (2.19), (2.20), and (2.21) that
I 0 = Im z 0 f z 0 f z 0 f z 0 z 0 - p - λ = - p β + λ β z 0 p z 0 .
(2.22)
In view of λ > 0, from (2.21) and (2.22) we obtain
I 0 - λ + λ + 2 p β 2 2 β λ λ + 2 p β < 0
(2.23)
and
I 0 - λ + λ + 2 p β 2 2 β - λ λ + 2 p β > 0 .
(2.24)

But both (2.23) and (2.24) contradict the assumption (2.17). Therefore, we have Rep(z) > 0 for all z U. This shows that (2.18) holds true.

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Suqian College
(2)
Department of Mathematics, Yangzhou University

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Copyright

© Xu et al.; licensee Springer. 2012

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